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Inessa05 [86]
2 years ago
5

There are slips of paper that are numbered from 1 to 15 in a hat. What is the theoretical probability that a slip of paper that

is chosen will be an even number?
1/15
7/15
1/2
8/15
Mathematics
2 answers:
blsea [12.9K]2 years ago
8 0
Okay! Let's write out all the numbers: 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Now, let's bold (or you can circle them... or do really anything to them as long as they are standing out to you) the even numbers (remember all even numbers are divisible by 2)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

Now let's list all the ones we highlighted: 
2
4
6
8
10
12
14

Count up how many numbers that is. 
You should get 7. 

So, our answer is 7 out of 15 numbers. 
Or as a fraction: 

7/15 is our answer.

Note: A probability is the number of chances something has to happen out of the total number of chances. So in this case the number of even numbers out of the total number of numbers. You can apply this to any probability problem. Hope this helps!

DochEvi [55]2 years ago
6 0
There are 7 even numbers between 1 and 15 so the answer will be 7/15

Hope this helps :)
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melamori03 [73]
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Since it is not that much, you would write it as g ≤ 3k - 3
6 0
2 years ago
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Anne has 24 more cards than Devi.  Anne finds that 3/5 of Devi's cards are equal to 1/2 of her cards.  How many cards does Anne
Sophie [7]
A=24xD
3/5xD=1/2xA
3d/5=12
d=12x5/3
d=20
a=480
8 0
1 year ago
This​ year, druehl,​ inc., will produce 57 comma 60057,600 hot water heaters at its plant in​ delaware, in order to meet expecte
ivolga24 [154]
If one labour works 200 hours per month, the amount of hot water heaters the labour can produce is = 200 × 0.25 = 50 hot water heaters

The demand is to produce 57600 hot water heaters

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5 0
2 years ago
A high school track is shaped as a rectangle with a half circle on either side.
earnstyle [38]

Answer:

Step-by-step explanation:

perimeter=2(85)+π(12.5)=220+2×12.5π

=170+25×3.14

=170+78.5

=248.5

distance ran=3×248.5=745.5 yards.

8 0
2 years ago
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Find the value of cosAcos2Acos3A...........cos998Acos999A where A=2π/1999
Lady bird [3.3K]
Hello,

Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.

Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)

As sin x *cos x=sin (2x) /2

P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
         *1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
 as sin(x)=-sin(2pi-x) and 2pi=1999a

sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)

So  sin(2a)*sin(4a)*...
     *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".

Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)

       








7 0
1 year ago
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